ECE 201, Section 3 Lecture 12. Prof. Peter Bermel September 17, 2012

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Transcription:

ECE 201, Section 3 Lecture 12 Prof. Peter ermel September 17, 2012

Exam #1: Thursday, Sep. 20 6:307:30 pm Most of you will be in WTHR 200, unless told otherwise Review session tonight at 8 pm (MTH 175) will go over posted Review Exam #1 Posted answers to 3 sample exams on lackboard to help you study I ll have office hours 12:301:30 pm MWF in EE 331 and by appointment 9/17/2012 ECE 2013, Prof. ermel

Exam #1 Key Concepts asic concepts: current, voltage, charge, Ohm s law, KCL, KVL, current & voltage division Combining resistors in series and parallel Resistor networks: node, supernode, loop analysis Source transformation 9/17/2012 ECE 2013, Prof. ermel

asic Concepts Current Irepresents the flow of charge: = Voltage V creates potential energy U for charges: = Power dissipated (passive sign convention): = Ohm s Law for resistors: = 9/17/2012 ECE 2013, Prof. ermel

asic Concepts: Kirchoff slaws Kirchoff s Current Law (KCL) Sum of all currents entering a node or Gaussian surface is zero at all times: =0, for all t Kirchoff s Voltage Law (KVL) Voltage drop between any two nodes is directiondependent and pathindependent (i.e., V =V V ) Sum of voltage drops over any closed loop is zero 8/29/2012 ECE 2013, Prof. ermel

Voltage and Current Division in Resistors; Resistor Networks Series resistors: = = / ; currents equal Parallel resistors = = / ; voltages equal Seriesparallel circuits nalyzed iteratively 9/17/2012 ECE 2013, Prof. ermel

Resistor Network nalysis pproaches Nodal analysis Modified nodal analysis Nodal analysis with floating voltage sources Loop analysis ll approaches should generally yield the same physical results est choice generally involves least number of unknowns, and will depend on details of problem 9/17/2012 ECE 2013, Prof. ermel

Formula Sheet for Exam 1 9/17/2012 ECE 2013, Prof. ermel

TestTaking Strategies Preparing for the exam: Review relevant notes (especially these) Practice exams (time at least one) Previous exams from this semester Read each question carefully and figure out exactly what to calculate; no trick questions Time management: No more than 4 minutes per question until all done Guess among plausible options, if needed Take 30 sec break if feeling stressed 9/17/2012 ECE 2013, Prof. ermel

Theveninand Norton Circuits Linear Passive Circuit V oc R th I sc R th Thévenin equivalent circuit Norton equivalent circuit Equivalent circuits related by: = 9/17/2012 ECE 2013, Prof. ermel

Example 1 For this network: What is the Thévenin equivalent circuit? What is the Norton equivalent circuit? 12 Ω 6 Ω 4 4 Ω 36 V 30 V 9/17/2012 ECE 2013, Prof. ermel

Solution Use source transformation theorem to write this equivalent circuit: 3 5 4 12 Ω 6 Ω 4 Ω 9/17/2012 ECE 2013, Prof. ermel

Solution Combine resistors and current sources in parallel to obtain Norton equivalent: 12 2 Ω Transform into Thévenin equivalent: 24 V 2 Ω 9/17/2012 ECE 2013, Prof. ermel

lternate Solution Use superposition to obtain: Open circuit voltage from individual contributions 12 Ω 6 Ω 0 12 Ω 6 Ω 0 12 Ω 6 Ω 4 4 Ω 4 Ω 4 Ω 36 V 0 V 0 V 30 V 0 V 0 V Δ = 12/5 12/512 36 Δ = 3 36 30 Δ =( 1 12 1 6 1 4 ) 4 V oc =6 10 8 V=24 V yields Théveninequivalent voltage 9/17/2012 ECE 2013, Prof. ermel

lternate Solution Use superposition to obtain: Short circuit current from individual contributions 12 Ω 6 Ω 0 12 Ω 6 Ω 0 12 Ω 6 Ω 4 4 Ω 4 Ω 4 Ω 36 V 0 V 0 V 30 V 0 V 0 V Δ = 1 12 Ω 36 Δ = 1 6 Ω 30 Δ =4 I sc =3 5 4 = 12 yields Norton equivalent current 9/17/2012 ECE 2013, Prof. ermel

lternate Solution Take quotient of Théveninvoltage and Norton current yields R th =2 Ωand the following equivalent circuits: 12 2 Ω 24 V 2 Ω 9/17/2012 ECE 2013, Prof. ermel

Equivalencies for Complex Circuits Thevenin theorem general 2terminal linear network obeying = must have R th =ρ, V oc =ν Norton theorem general 2terminal linear network following = must have G th =γ, I sc =σ 9/17/2012 ECE 2013, Prof. ermel

Example 2 (a) Find the Thévenin equivalent of the circuit below (b) Find the output current and power at the load 24 kω 66 kω I 2 kω 36 kω 9 kω 2.5 m 30 V 9 kω 18 kω 18 V 9/17/2012 ECE 2013, Prof. ermel

Solution 24 kω 66 kω I 2 kω 36 kω 9 kω 2.5 m 30 V 9 kω 18 kω 18 V = =6 kω; =1m; V=6 V 9/17/2012 ECE 2013, Prof. ermel

Solution 24 kω I 66 kω I C I 2 kω 36 kω 9 kω 2.5 m 30 V 6 kω 6 V = =24 kω =2.5m; 36 = ; 666 =30 6 9/17/2012 ECE 2013, Prof. ermel

Solution I 2 kω 24 kω 52 V = =24 kω =13/9m; =52 ; =2m; =8mW 9/17/2012 ECE 2013, Prof. ermel

Homework HW #11 due today by 4:30 pm in EE 325 HW #12 due Fri.: DeCarlo& Lin, Chapter 6: Problem 3 Problem 6 Problem 8(a),(b) 9/17/2012 ECE 2013, Prof. ermel

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